题意：给你一个数字，转换成罗马数字，需要遵循罗马数字排布的一些规则。
解题思路：
思路一：因为9和4比较特殊，其余的就是直接累计，可以把罗马数字分成9,5,4,1的对应表，然后逐一遍历将对应的罗马符号放入该位即可。
class Solution {
public:
string intToRoman(int num) {
string ans;
vector<vector<string>> ivec;
vector<string> ivec1 = {"IX","V","IV","I"};
vector<string> ivec2 = {"XC","L","XL","X"};
vector<string> ivec3 = {"CM","D","CD","C"};
vector<string> ivec4 = {" "," "," ","M"};
ivec.push_back(ivec1);
ivec.push_back(ivec2);
ivec.push_back(ivec3);
ivec.push_back(ivec4);
int a[4] = {9, 5, 4, 1};
int bt = 0;
while(num){
int n = num % 10;
string ss;
while(n){
for(int i = 0; i < 4; i++){
//cout << a[i] << endl;
if(n >= a[i]){
ss += ivec[bt][i];
n -= a[i];
}
}
}
ans = ss + ans;
num /= 10;
bt++;
}
return ans;
}
};
思路二：由于题目给出条件，说数字不会超过3999，直接暴力更快更准确。
class Solution {
public:
string intToRoman(int num) {
string a[10] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
string b[10] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
string c[10] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
string d[10] = {"", "M", "MM", "MMM", "", "", "", "", "", ""};
return d[num/1000] + c[(num/100)%10] + b[(num/10)%10] + a[num%10];
}
};
真 * 牛逼！
